javascript - 根据多个属性在数组中查找唯一对象

Question

我需要根据下面的 2 个属性从数组中找到唯一的对象。当“class”和“fare”匹配时，我需要提取唯一值并将它们放入结果数组中。

资源：

 var arr = [{class:"second", fare: "a"},
 {class:"second", fare: "b"},
 {class:"first", fare: "a"},
 {class:"first", fare: "a"},
 {class:"second", fare: "a"},
 {class:"first", fare: "c"}
]

预期结果：

 var result = [{class:"second", fare: "a"},
 {class:"second", fare: "b"},
 {class:"first", fare: "a"},
 {class:"first", fare: "c"}
]

我查看了 SO 并找到了基于一个属性过滤的答案（ Create array of unique objects by property ），但找不到基于 2 个属性的答案。

原文由 whyAto8 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

您可以为哈希表构建组合键并过滤给定的数组。

 var arr = [{ class: "second", fare: "a" }, { class: "second", fare: "b" }, { class: "first", fare: "a" }, { class: "first", fare: "a" }, { class: "second", fare: "a" }, { class: "first", fare: "c" }],
    result = arr.filter(function (a) {
        var key = a.class + '|' + a.fare;
        if (!this[key]) {
            this[key] = true;
            return true;
        }
    }, Object.create(null));

console.log(result);

 .as-console-wrapper { max-height: 100% !important; top: 0; }

同样没有（误）使用 thisArg 的 Array#filter 。

 var array = [{ class: "second", fare: "a" }, { class: "second", fare: "b" }, { class: "first", fare: "a" }, { class: "first", fare: "a" }, { class: "second", fare: "a" }, { class: "first", fare: "c" }],
    seen = Object.create(null),
    result = array.filter(o => {
        var key = ['class', 'fare'].map(k => o[k]).join('|');
        if (!seen[key]) {
            seen[key] = true;
            return true;
        }
    });

console.log(result);

 .as-console-wrapper { max-height: 100% !important; top: 0; }

原文由 Nina Scholz 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 2

使用组合键和地图

 const array = [
 { class: "second", fare: "a" },
 { class: "second", fare: "b" },
 { class: "first", fare: "a" },
 { class: "first", fare: "a" },
 { class: "second", fare: "a" },
 { class: "first", fare: "c" }
];

console.log(unique(array, ['class', 'fare']));

function unique(arr, keyProps) {
 const kvArray = arr.map(entry => {
  const key = keyProps.map(k => entry[k]).join('|');
  return [key, entry];
 });
 const map = new Map(kvArray);
 return Array.from(map.values());
}

原文由 Tamo Maes 发布，翻译遵循 CC BY-SA 3.0 许可协议